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identify the identity: (1-sin theta) / (1+sin theta) = (tan theta -sec theta)^2
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If you don't mind, I'll forgo the writing of \(\theta\) because it's a bit tedious. \[\begin{align*}\frac{1-\sin}{1+\sin}&=\frac{(1-\sin)(1-\sin)}{(1+\sin)(1-\sin)}\\ &=\frac{(1-\sin)^2}{1-\sin^2}\\ &=\frac{1-2\sin+\sin^2}{\cos^2}\\ &=\frac{1}{\cos^2}-2\frac{\sin}{\cos^2}+\frac{\sin^2}{\cos^2}\\ &=\sec^2-2\tan\sec+\tan^2\\ &=(\sec-\tan)^2\\ &=((-1)(\tan-\sec))^2\\ &=(-1)^2(\tan-\sec)^2\\ &=(\tan-\sec)^2\end{align*}\]
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